AMC 10 · 2017 · #18

Grade 8 probability
Archetype Branching State Enumeration └ Sub-archetype Markov State Probability
probability-basicsequences-geometric convert-to-algebra ↑ Prerequisites: probability-basic
📏 Medium solution 💡 2 insights
Problem
Amelia and Blaine take turns tossing their own coins, Amelia first, and the first person to toss a head wins. Amelia's coin shows heads with probability 1/3 and Blaine's with probability 2/5, and all tosses are independent. If the probability that Amelia wins is the reduced fraction p/q, find q-p.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Tool #4 (Introduce a Variable): the game could last forever, so instead of adding up infinitely many turns I name the answer a = the probability Amelia wins and use the game's self-similarity. The key observation is that after Amelia misses and then Blaine misses, it is Amelia's turn again with the exact same situation, so the chance of winning from there is again a. Tool #16 (Change Focus / Count the Complement): the only way the game returns to Amelia is that nobody wins this round, i.e. both players get tails, which is the complement of an immediate win on each toss. Tool #13 (Convert to Algebra): the self-similarity turns into a single linear equation in a that I can solve directly, avoiding the infinite series entirely.

1STEP 1

Name the answer and spot the repeat

Let a be Amelia's chance of winning measured from her own turn; if both miss a full round, it is her turn again with that same chance a.

a = Pr(Amelia wins from her turn)
2STEP 2

Chance the game comes back to Amelia

No winner this round means Amelia tails (2/3) and Blaine tails (3/5); independence lets us multiply, giving 2/5.

2/3·3/5=6/15=2/5
3STEP 3

Build and solve the equation

Either Amelia wins now (1/3) or the round returns (2/5) and she wins with a again, so a=1/3+2/5·a, giving a=5/9.

a=1/3+2/5a → 3/5a=1/3 → a=1/3·5/3=5/9
4STEP 4

Read off p, q and subtract

Since gcd(5,9)=1, the fraction 5/9 is already reduced, so p=5, q=9 and q-p=9-5=4, choice (D).

gcd(5,9)=1 → p=5, q=9 → q-p=9-5=4 → (D)
Answer
4
Going first is an advantage, so Amelia should win more than half the time, and 5/9≈ 0.556 is indeed just above 1/2 — sensible since her per-toss head chance 1/3 is a bit lower than Blaine's 2/5, but moving first roughly offsets it. A quick check of the first two terms of the full series, 1/3+2/5·1/3=1/3+2/15=7/15≈ 0.467, is already climbing toward 5/9, confirming the equation's value. Then q-p=4 is among the choices, so (D) fits.
💡Key takeaway

If a game can repeat itself forever, name the answer a, notice the leftover game is a copy of the start, write one equation like a=1/3+2/5a, and solve.

  • Name the answer and spot the repeat
  • Chance the game comes back to Amelia
  • Build and solve the equation
  • Read off p, q and subtract

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